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hram777 [196]
2 years ago
10

What is the purpose of the zigzag line on the right side of the periodic table?

Chemistry
1 answer:
Wittaler [7]2 years ago
5 0

Answer:

D.

Explanation:

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What is the pOH of a<br> 5.6 x 10-5 M solution of cesium<br> hydroxide (CsOH)?
Rudiy27

Answer: 4.25

Explanation: CsOH=Cs+

pOH= -log10

= -log10(5.6*10^-5)

= -log10(5.6*10^-5)=-(-4.25)

=4.25

7 0
3 years ago
SnO2 + 2 H2 ——&gt; Sn + 2 H2O
SpyIntel [72]

Answer:

0.15g

Explanation:

Given parameters:

Number of molecules of water = 1.2 x 10²¹ molecules

Unknown:

Mass of SnO₂  = ?

Solution:

To solve this problem, we have to work from the known to the unknown specie;

             SnO₂   +    2H₂    →   Sn  +   2H₂O

Ensure that the equation given is balanced;

       

Now,

          the known species is water;

                  6.02 x 10²³ molecules of water  = 1 mole

                   1.2 x 10²¹ molecules of water  = \frac{1.2 x 10^{21} }{6.02 x 10^{23} }    = 0.2 x 10⁻²moles

Number of moles of water  = 0.002moles

           From the balanced chemical equation:

         

             2 mole of water is produced from 1 mole of    SnO₂  

           0.002 moles of water will be produced from \frac{0.002}{2}  = 0.001moles

To find the mass;

           Mass  = number of moles x molar mass

Molar mass of  SnO₂ = 118.7 + 2(16) = 150.7g/mol

        Mass  =  0.001 x 150.7 = 0.15g

3 0
3 years ago
Which of these is NOT a layer of the skin?
bagirrra123 [75]
I need the options to choose from
4 0
2 years ago
Read 2 more answers
Given the following data:
bagirrra123 [75]

176.0 \; \text{kJ} \cdot \text{mol}^{-1}

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its \Delta H can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with \Delta H given be denoted as (1), (2), (3), and the last equation (4). Let a, b, and c be letters such that a \times (1) + b \times (2) + c \times (3) = (4). This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 3 + (-1) = 2 shall resemble the number of \text{H}_2 left on the product side when the second equation is directly added to the third. Similarly

  • \text{NH}_4 \text{Cl} \; (s): -2 \; a = 1
  • \text{NH}_3\; (g): -2 \; b = -1
  • \text{HCl} \; (g): 2 \; c = -1

Thus

a = -1/2\\b = 1/2\\c = -1/2 and

-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)

Verify this conclusion against a fourth species involved- \text{N}_2 \; (g) for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

a + b = -1/2 + 1/2 = 0

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}

3 0
3 years ago
Please show all work for full credit.
koban [17]

Answer:

No, it is not sufficient

Please find the workings below

Explanation:

Using E = hf

Where;

E = energy of a photon (J)

h = Planck's constant (6.626 × 10^-34 J/s)

f = frequency

However, λ = v/f

f = v/λ

Where; λ = wavelength of light = 325nm = 325 × 10^-9m

v = speed of light (3 × 10^8 m/s)

Hence, E = hv/λ

E = 6.626 × 10^-34 × 3 × 10^8 ÷ 325 × 10^-9

E = 19.878 × 10^-26 ÷ 325 × 10^-9

E = 19.878/325 × 10^ (-26+9)

E = 0.061 × 10^-17

E = 6.1 × 10^-19J

Next, we work out the energy required to dissociate 1 mole of N=N. Since the bond energy is 418 kJ/mol.

E = 418 × 10³ ÷ 6.022 × 10^23

E = 69.412 × 10^(3-23)

E = 69.412 × 10^-20

E = 6.9412 × 10^-19J

6.9412 × 10^-19J is required to break one mole of N=N bond.

Based on the workings above, the photon, which has an energy of 6.1 × 10^-19J is not sufficient to break a N=N bond that has an energy of 6.9412 × 10^-19J

8 0
3 years ago
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